1. Determine the direction of the missing vector, U, B, or F, in the scenarios shown in the figure below. All moving charges are positive. You don't need to show any work or explain your answer, just tell me the missing vector and where it points. (a) 15 (b) + (c) 15 →F 14 F 8 15 (d) OF OF (e) 150 B (f) OB F | 14 F

Question 14 of 35 A particle that carries a net charge of-41.8 uC is held in a constant electric field that is uniform over the entire region. The electric field vector is oriented 25.2° clockwise from the vertical axis, as shown in the figure. If the magnitude of the electric field is 4.32 N/C, how much work is done by the electric field as the particle is made to E move a distance of d = 0.556 m straight up? d. 25.2° work: J What is the potential difference AV between the particle's initial and final positions? AV =

Ch 18 q3 In 1910 Rutherford performed a classic experiment in which he directed a beam of alpha particles at a thin gold foil. He unexpectedly observed a few of the particles scattered almost directly backward. This result was not consistent with then current models of atomic structure and led Rutherford to propose the existence of a very dense concentration of positive charge at the center of an atom—the atomic nucleus. The alpha particle has a charge of +2e and the gold nucleus a charge of +79e. Suppose that an alpha particle is initially a great distance from the gold, has a kinetic energy of 4.20 MeV (4.20  106 eV), and is headed directly at a gold nucleus. How close will the particle come to the center of the nucleus? Treat the nucleus and the alpha particle as point charges.

A uniform electric field with magnitude of 8.00 * 104 that is directed vertically upward has a m charge of 40.0 nC placed inside. What is the change in potential energy of moving the charge a. 0.65 m to the right b. 0.85 m vertically upward C. 0.40 m downward at an angle of 25° from the horizontal